If $\sum x_n$ converges, is the even partial sums of the sequence squared Cauchy?

If $$\sum x_n$$ converges, is the sequence $$a_k =\sum_{n=1}^k x_{2n}^2$$ Cauchy?

I suspect that it isn't but a bit stuck on finding a counter example. For a valid counterexample, I believe I need a series that converges but the squared even partial sum does not.

Writing the definition of Cauchy, we know a sequence $$(x_n)$$ is Cauchy if for every $$\epsilon > 0$$, there exists $$N \in \mathbb{N}$$ such that $$m, n \geq N \implies |x_m -x_n| < \epsilon$$. But it seems easier to just use that convergence implies Cauchy rather than work directly from the definition. Any ideas/tips on what exactly I'm not seeing?

• You cannot find a counter-example if the sequence is nonnegative since then $x_n^2 \leq x_n$ for all sufficiently large $n$. So... – Michael Oct 15 '18 at 1:35
• If $x_n=(-1)^n/\sqrt n$ then $\sum_n x_n$ converges but $\sum_n x_n^2$ diverges. (Yes, I am aware that this isn't your exact problem.) – Lord Shark the Unknown Oct 15 '18 at 1:35
• @Michael What if $x_n<0$? – Lord Shark the Unknown Oct 15 '18 at 1:37
• @LordSharktheUnknown : I don't follow your comment directed to me ("@Michael what if..."), I think you meant something else or you misread my first comment. Your first comment seems in teh same spirit as mine but gives a bit of a larger hint. – Michael Oct 15 '18 at 1:38
• I see now there is a third hinter. With all these "hints" the problem will soon be solved without the asker's involvement! – Michael Oct 15 '18 at 1:45

Hint: Let $$\{y_n\}$$ be a sequence of positive reals that converges to $$0$$, and take

$$x_n=\begin{cases} -y_{n/2} & \mathrm{if\ }n\equiv 0\bmod 2 \\ y_{(n+1)/2} & \mathrm{if\ }n\equiv 1\bmod 2,\end{cases}$$

or in other words $$x_1,x_2,\cdots,$$ is $$y_1,-y_1,y_2,-y_2,y_3,-y_3\cdots$$.

$$\sum_{n=1}^k y_n^2.$$
Can you find a sequence $$\{y_n\}$$ that converges to $$0$$ for which the above does not converge?
• This doesn't converge to 0, but it seems this would work for the question, $\sum \frac{(-1)^{n+1}}{n}$? Converges but the above squared doesn't converge. – SS' Oct 15 '18 at 2:12
• @SS' Isn't the square of it $\sum \frac{1}{n^2}$, which converges? However, that idea almost works - what if you put a slower-growing function in the denominator? – Carl Schildkraut Oct 15 '18 at 2:46